Problem: Multiply the following complex numbers: $({3-2i}) \cdot ({-2-2i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({3-2i}) \cdot ({-2-2i}) = $ $ ({3} \cdot {-2}) + ({3} \cdot {-2}i) + ({-2}i \cdot {-2}) + ({-2}i \cdot {-2}i) $ Then simplify the terms: $ (-6) + (-6i) + (4i) + (4 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -6 + (-6 + 4)i + 4i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -6 + (-6 + 4)i - 4 $ The result is simplified: $ (-6 - 4) + (-2i) = -10-2i $